{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "google",
   "metadata": {},
   "source": [
    "##### Copyright 2025 Google LLC."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "apache",
   "metadata": {},
   "source": [
    "Licensed under the Apache License, Version 2.0 (the \"License\");\n",
    "you may not use this file except in compliance with the License.\n",
    "You may obtain a copy of the License at\n",
    "\n",
    "    http://www.apache.org/licenses/LICENSE-2.0\n",
    "\n",
    "Unless required by applicable law or agreed to in writing, software\n",
    "distributed under the License is distributed on an \"AS IS\" BASIS,\n",
    "WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.\n",
    "See the License for the specific language governing permissions and\n",
    "limitations under the License.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "basename",
   "metadata": {},
   "source": [
    "# nonogram_table"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "link",
   "metadata": {},
   "source": [
    "<table align=\"left\">\n",
    "<td>\n",
    "<a href=\"https://colab.research.google.com/github/google/or-tools/blob/main/examples/notebook/contrib/nonogram_table.ipynb\"><img src=\"https://raw.githubusercontent.com/google/or-tools/main/tools/colab_32px.png\"/>Run in Google Colab</a>\n",
    "</td>\n",
    "<td>\n",
    "<a href=\"https://github.com/google/or-tools/blob/main/examples/contrib/nonogram_table.py\"><img src=\"https://raw.githubusercontent.com/google/or-tools/main/tools/github_32px.png\"/>View source on GitHub</a>\n",
    "</td>\n",
    "</table>"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "doc",
   "metadata": {},
   "source": [
    "First, you must install [ortools](https://pypi.org/project/ortools/) package in this colab."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "install",
   "metadata": {},
   "outputs": [],
   "source": [
    "%pip install ortools"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "description",
   "metadata": {},
   "source": [
    "\n",
    "\n",
    "  Nonogram  (Painting by numbers) in Google CP Solver.\n",
    "\n",
    "  http://en.wikipedia.org/wiki/Nonogram\n",
    "  '''\n",
    "  Nonograms or Paint by Numbers are picture logic puzzles in which cells in a\n",
    "  grid have to be colored or left blank according to numbers given at the\n",
    "  side of the grid to reveal a hidden picture. In this puzzle type, the\n",
    "  numbers measure how many unbroken lines of filled-in squares there are\n",
    "  in any given row or column. For example, a clue of '4 8 3' would mean\n",
    "  there are sets of four, eight, and three filled squares, in that order,\n",
    "  with at least one blank square between successive groups.\n",
    "\n",
    "  '''\n",
    "\n",
    "  See problem 12 at http://www.csplib.org/.\n",
    "\n",
    "  http://www.puzzlemuseum.com/nonogram.htm\n",
    "\n",
    "  Haskell solution:\n",
    "  http://twan.home.fmf.nl/blog/haskell/Nonograms.details\n",
    "\n",
    "  Brunetti, Sara & Daurat, Alain (2003)\n",
    "  'An algorithm reconstructing convex lattice sets'\n",
    "  http://geodisi.u-strasbg.fr/~daurat/papiers/tomoqconv.pdf\n",
    "\n",
    "\n",
    "  The Comet model (http://www.hakank.org/comet/nonogram_regular.co)\n",
    "  was a major influence when writing this Google CP solver model.\n",
    "\n",
    "  I have also blogged about the development of a Nonogram solver in Comet\n",
    "  using the regular constraint.\n",
    "  * 'Comet: Nonogram improved: solving problem P200 from 1:30 minutes\n",
    "     to about 1 second'\n",
    "     http://www.hakank.org/constraint_programming_blog/2009/03/comet_nonogram_improved_solvin_1.html\n",
    "\n",
    "  * 'Comet: regular constraint, a much faster Nonogram with the regular\n",
    "  constraint,\n",
    "     some OPL models, and more'\n",
    "     http://www.hakank.org/constraint_programming_blog/2009/02/comet_regular_constraint_a_muc_1.html\n",
    "\n",
    "  Compare with the other models:\n",
    "  * Gecode/R: http://www.hakank.org/gecode_r/nonogram.rb (using 'regexps')\n",
    "  * MiniZinc: http://www.hakank.org/minizinc/nonogram_regular.mzn\n",
    "  * MiniZinc: http://www.hakank.org/minizinc/nonogram_create_automaton.mzn\n",
    "  * MiniZinc: http://www.hakank.org/minizinc/nonogram_create_automaton2.mzn\n",
    "    Note: nonogram_create_automaton2.mzn is the preferred model\n",
    "\n",
    "  This model was created by Hakan Kjellerstrand (hakank@gmail.com)\n",
    "  Also see my other Google CP Solver models:\n",
    "  http://www.hakank.org/google_or_tools/\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "code",
   "metadata": {},
   "outputs": [],
   "source": [
    "import sys\n",
    "\n",
    "from ortools.constraint_solver import pywrapcp\n",
    "\n",
    "\n",
    "#\n",
    "# Global constraint regular\n",
    "#\n",
    "# This is a translation of MiniZinc's regular constraint (defined in\n",
    "# lib/zinc/globals.mzn), via the Comet code refered above.\n",
    "# All comments are from the MiniZinc code.\n",
    "# '''\n",
    "# The sequence of values in array 'x' (which must all be in the range 1..S)\n",
    "# is accepted by the DFA of 'Q' states with input 1..S and transition\n",
    "# function 'd' (which maps (1..Q, 1..S) -> 0..Q)) and initial state 'q0'\n",
    "# (which must be in 1..Q) and accepting states 'F' (which all must be in\n",
    "# 1..Q).  We reserve state 0 to be an always failing state.\n",
    "# '''\n",
    "#\n",
    "# x : IntVar array\n",
    "# Q : number of states\n",
    "# S : input_max\n",
    "# d : transition matrix\n",
    "# q0: initial state\n",
    "# F : accepting states\n",
    "def regular(x, Q, S, d, q0, F):\n",
    "\n",
    "  solver = x[0].solver()\n",
    "\n",
    "  assert Q > 0, 'regular: \"Q\" must be greater than zero'\n",
    "  assert S > 0, 'regular: \"S\" must be greater than zero'\n",
    "\n",
    "  # d2 is the same as d, except we add one extra transition for\n",
    "  # each possible input;  each extra transition is from state zero\n",
    "  # to state zero.  This allows us to continue even if we hit a\n",
    "  # non-accepted input.\n",
    "\n",
    "  d2 = []\n",
    "  for i in range(Q + 1):\n",
    "    for j in range(S):\n",
    "      if i == 0:\n",
    "        d2.append((0, j, 0))\n",
    "      else:\n",
    "        d2.append((i, j, d[i - 1][j]))\n",
    "\n",
    "  # If x has index set m..n, then a[m-1] holds the initial state\n",
    "  # (q0), and a[i+1] holds the state we're in after processing\n",
    "  # x[i].  If a[n] is in F, then we succeed (ie. accept the\n",
    "  # string).\n",
    "  x_range = list(range(0, len(x)))\n",
    "  m = 0\n",
    "  n = len(x)\n",
    "\n",
    "  a = [solver.IntVar(0, Q + 1, 'a[%i]' % i) for i in range(m, n + 1)]\n",
    "\n",
    "  # Check that the final state is in F\n",
    "  solver.Add(solver.MemberCt(a[-1], F))\n",
    "  # First state is q0\n",
    "  solver.Add(a[m] == q0)\n",
    "  for i in x_range:\n",
    "    solver.Add(x[i] >= 1)\n",
    "    solver.Add(x[i] <= S)\n",
    "    # Determine a[i+1]: a[i+1] == d2[a[i], x[i]]\n",
    "    solver.Add(solver.AllowedAssignments((a[i], x[i] - 1, a[i + 1]), d2))\n",
    "\n",
    "\n",
    "#\n",
    "# Make a transition (automaton) matrix from a\n",
    "# single pattern, e.g. [3,2,1]\n",
    "#\n",
    "\n",
    "\n",
    "def make_transition_matrix(pattern):\n",
    "\n",
    "  p_len = len(pattern)\n",
    "  num_states = p_len + sum(pattern)\n",
    "\n",
    "  # this is for handling 0-clues. It generates\n",
    "  # just the state 1,2\n",
    "  if num_states == 0:\n",
    "    num_states = 1\n",
    "\n",
    "  t_matrix = []\n",
    "  for i in range(num_states):\n",
    "    row = []\n",
    "    for j in range(2):\n",
    "      row.append(0)\n",
    "    t_matrix.append(row)\n",
    "\n",
    "  # convert pattern to a 0/1 pattern for easy handling of\n",
    "  # the states\n",
    "  tmp = [0 for i in range(num_states)]\n",
    "  c = 0\n",
    "  tmp[c] = 0\n",
    "  for i in range(p_len):\n",
    "    for j in range(pattern[i]):\n",
    "      c += 1\n",
    "      tmp[c] = 1\n",
    "    if c < num_states - 1:\n",
    "      c += 1\n",
    "      tmp[c] = 0\n",
    "\n",
    "  t_matrix[num_states - 1][0] = num_states\n",
    "  t_matrix[num_states - 1][1] = 0\n",
    "\n",
    "  for i in range(num_states):\n",
    "    if tmp[i] == 0:\n",
    "      t_matrix[i][0] = i + 1\n",
    "      t_matrix[i][1] = i + 2\n",
    "    else:\n",
    "      if i < num_states - 1:\n",
    "        if tmp[i + 1] == 1:\n",
    "          t_matrix[i][0] = 0\n",
    "          t_matrix[i][1] = i + 2\n",
    "        else:\n",
    "          t_matrix[i][0] = i + 2\n",
    "          t_matrix[i][1] = 0\n",
    "\n",
    "  # print 'The states:'\n",
    "  # for i in range(num_states):\n",
    "  #     for j in range(2):\n",
    "  #         print t_matrix[i][j],\n",
    "  #     print\n",
    "  # print\n",
    "\n",
    "  return t_matrix\n",
    "\n",
    "\n",
    "#\n",
    "# check each rule by creating an automaton\n",
    "# and regular\n",
    "#\n",
    "\n",
    "\n",
    "def check_rule(rules, y):\n",
    "  solver = y[0].solver()\n",
    "\n",
    "  r_len = sum([1 for i in range(len(rules)) if rules[i] > 0])\n",
    "  rules_tmp = []\n",
    "  for i in range(len(rules)):\n",
    "    if rules[i] > 0:\n",
    "      rules_tmp.append(rules[i])\n",
    "\n",
    "  transition_fn = make_transition_matrix(rules_tmp)\n",
    "  n_states = len(transition_fn)\n",
    "  input_max = 2\n",
    "\n",
    "  # Note: we cannot use 0 since it's the failing state\n",
    "  initial_state = 1\n",
    "  accepting_states = [n_states]  # This is the last state\n",
    "\n",
    "  regular(y, n_states, input_max, transition_fn, initial_state,\n",
    "          accepting_states)\n",
    "\n",
    "\n",
    "def main(rows, row_rule_len, row_rules, cols, col_rule_len, col_rules):\n",
    "\n",
    "  # Create the solver.\n",
    "  solver = pywrapcp.Solver('Regular test')\n",
    "\n",
    "  #\n",
    "  # data\n",
    "  #\n",
    "\n",
    "  #\n",
    "  # variables\n",
    "  #\n",
    "  board = {}\n",
    "  for i in range(rows):\n",
    "    for j in range(cols):\n",
    "      board[i, j] = solver.IntVar(1, 2, 'board[%i, %i]' % (i, j))\n",
    "  board_flat = [board[i, j] for i in range(rows) for j in range(cols)]\n",
    "\n",
    "  # Flattened board for labeling.\n",
    "  # This labeling was inspired by a suggestion from\n",
    "  # Pascal Van Hentenryck about my Comet nonogram model.\n",
    "  board_label = []\n",
    "  if rows * row_rule_len < cols * col_rule_len:\n",
    "    for i in range(rows):\n",
    "      for j in range(cols):\n",
    "        board_label.append(board[i, j])\n",
    "  else:\n",
    "    for j in range(cols):\n",
    "      for i in range(rows):\n",
    "        board_label.append(board[i, j])\n",
    "\n",
    "  #\n",
    "  # constraints\n",
    "  #\n",
    "  for i in range(rows):\n",
    "    check_rule([row_rules[i][j] for j in range(row_rule_len)],\n",
    "               [board[i, j] for j in range(cols)])\n",
    "\n",
    "  for j in range(cols):\n",
    "    check_rule([col_rules[j][k] for k in range(col_rule_len)],\n",
    "               [board[i, j] for i in range(rows)])\n",
    "\n",
    "  #\n",
    "  # solution and search\n",
    "  #\n",
    "  db = solver.Phase(board_label, solver.CHOOSE_FIRST_UNBOUND,\n",
    "                    solver.ASSIGN_MIN_VALUE)\n",
    "\n",
    "  solver.NewSearch(db)\n",
    "\n",
    "  num_solutions = 0\n",
    "  while solver.NextSolution():\n",
    "    print()\n",
    "    num_solutions += 1\n",
    "    for i in range(rows):\n",
    "      row = [board[i, j].Value() - 1 for j in range(cols)]\n",
    "      row_pres = []\n",
    "      for j in row:\n",
    "        if j == 1:\n",
    "          row_pres.append('#')\n",
    "        else:\n",
    "          row_pres.append(' ')\n",
    "      print('  ', ''.join(row_pres))\n",
    "\n",
    "    print()\n",
    "    print('  ', '-' * cols)\n",
    "\n",
    "    if num_solutions >= 2:\n",
    "      print('2 solutions is enough...')\n",
    "      break\n",
    "\n",
    "  solver.EndSearch()\n",
    "  print()\n",
    "  print('num_solutions:', num_solutions)\n",
    "  print('failures:', solver.Failures())\n",
    "  print('branches:', solver.Branches())\n",
    "  print('WallTime:', solver.WallTime(), 'ms')\n",
    "\n",
    "\n",
    "#\n",
    "# Default problem\n",
    "#\n",
    "# From http://twan.home.fmf.nl/blog/haskell/Nonograms.details\n",
    "# The lambda picture\n",
    "#\n",
    "rows = 12\n",
    "row_rule_len = 3\n",
    "row_rules = [[0, 0, 2], [0, 1, 2], [0, 1, 1], [0, 0, 2], [0, 0, 1], [0, 0, 3],\n",
    "             [0, 0, 3], [0, 2, 2], [0, 2, 1], [2, 2, 1], [0, 2, 3], [0, 2, 2]]\n",
    "\n",
    "cols = 10\n",
    "col_rule_len = 2\n",
    "col_rules = [[2, 1], [1, 3], [2, 4], [3, 4], [0, 4], [0, 3], [0, 3], [0, 3],\n",
    "             [0, 2], [0, 2]]\n",
    "\n",
    "if len(sys.argv) > 1:\n",
    "  file = sys.argv[1]\n",
    "  exec(compile(open(file).read(), file, 'exec'))\n",
    "main(rows, row_rule_len, row_rules, cols, col_rule_len, col_rules)\n",
    "\n"
   ]
  }
 ],
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